Re: ZFC++

On Mar 1, 9:48 pm, "Rupert" <rupertmccal...@xxxxxxxxx> wrote:
On Mar 2, 12:06 pm, "zuhair" <zaljo...@xxxxxxxxx> wrote:





On Mar 1, 6:29 pm, "Rupert" <rupertmccal...@xxxxxxxxx> wrote:

On Mar 2, 4:14 am, "zuhair" <zaljo...@xxxxxxxxx> wrote:

On Mar 1, 1:57 am, "Rupert" <rupertmccal...@xxxxxxxxx> wrote:

On Mar 1, 2:21 pm, "zuhair" <zaljo...@xxxxxxxxx> wrote:

Hi all,

ZFC+Proper classes+Ur elements'ZFC++':is the set of sentences entailed
(from first order logic with identity)by these axioms:

1) Extensionality: AxAy(x=y<->Az(zex<->zey)).

2) ur-element:Ex(Ay(yex<->y=x)).

Definition 1) x is a ur-element<-> Ay(yex<->y=x).

3) Regularity:

Ax((Ez(zex))->Ey(yex & ~Ec(cey & cex & ~c=y))).

4) Schema of Global comprehension: if P is a formula in which x is not
free then all closures of:

ExAy(yex<->(P(y)&Ez(yez)))

are axioms.

Definition 2) x=V <-> Ay(yex<->(y=y & Ez(yez))).

Definition 3) x is a proper class <-> ~Ey(xey).

Accordingly V is a proper class.

Definition 4) x=0 <-> Ay(~yex).

5) Pairing:AaeVAbeVExeVAyeV(yex<->(y=a or y=b)).

6) Union:AaExAy((yex<->Ez(zea&yez))&(Em(aem)<->En(xen))).

7) Infinity:ENeV(0eN&(Ax(xeN->xU{x}eN))).

with 'U' and {x} having the usual definitions.

8) limitation of size:

Ax((Ez(xez)) <-> x is subnumerous to V).

x is subnumerous to V <-> Af((f:x->V) ->(f is injective & ~ f is
surjective)).

9)Power: AaExAyeV((yex<->Az(zey->zea)&(Em(aem)<->En(xen)).

Definition 5) x=K<-> Ay(yex<->y is a ur-element)

so K has all ur-elements as its members.

10)ur-multiplicity:

Ax((Az(zex->zeK)&Eu(xeu))->Ey(y is a ur-element & ~yex)).

Accordingly K is a proper class.

/

Zuhair

It's not quite correct to call this ZFC++, because the axiom of
replacement cannot be proved. This theory actually has the same
consistency strength as Z. It is equi-interpretable with a theory
which stands in the same relation to Z as NBG does to ZF.

Actually I myself do not know if this theory is consistent or not, or
what theories it is equi-interpretable with.
But regarding your last statement:'replacement cannot be proved in
this theory', I want to know why?
Since this theory has axiom of limitation of size, and it has
global comprehension, then I thought replacement can be proved in this
theorum , in the same manner as how it is proved in NBG.Mind you that
replacment here means that one which is equivalent to that in ZFC, i.e
replacement of small sets: (AxeV(x is not a ur-element)E!yeV(y is not
a ur-element):P(x,y))->AaeV(a is not a ur-element)EbeV(b is not a ur-
element)AyeV(y is not a ur-element)(yeb<->Ex(xea):P(x,y)).

Actually we can have a replacement containing ur-elements also.(AxeVE!
yeV:P(x,y))->AaeVEbeVAyeV(yeb<->Ex(xea):P(x,y)).

we can even have larger replacement:
(AxE!yeV:P(x,y))->AaEbAyeV(yeb<->Ex(xea):P(x,y)).

I think all of these can be proved in this theory.

Well, I don't think so. Can you elaborate on exactly how you would go
about proving them?

Same thing can be said of separation.

I think separation is provable, yes.

So I think you might have made a hasty remark???

Well, it's possible, but my proof that the axiom of replacement is not
provable looks fairly sound to me.

Oh! you made one!? I must have overlooked it then.
I didn't see this proof. You didn't wright it.
You only mentioned it.

I should see this proof.

Frankly speaking I think you can't bring such a proof, because A
subtheory of this theory is equivalent to Morse-Kelley. That's why I
say you must be wrong, though I didn't see this proof of yours yet.

I would like to see this proof, can you post it, or at least tell me
where you've post it.

Zuhair

Let S be a set of cardinality beth-(omega-times-two).

what is beth-(Omega-times-two).

Call the
elements of S the "ur-elements".

This is vague to me what do you mean by this. Is S a subclass of the
class of all ur-elements or Is S is the class of all ur-elements
itself. I would guess that beth-(Omega-Times-Two) is not the biggest
cardinal. So S should be a subclass of the class of all ur-elements.


Let V_0=S. For each ordinal alpha
such that 0<alpha<(omega times 2)+1, if alpha=beta+1 let V_alpha be
the set of all subsets of V_beta with cardinality less than beth-omega-
times-2, if alpha is a limit ordinal let V_alpha be the union of all
V_beta with beta<alpha. The union of the V's is a model for your
theory.

what do you mean a model in this theory.is it a small set
or a large set. I think you mean it is a large set.


This yields an interpretation of your theory in a theory which
stands to Z as NBG stands to ZF. It is easy to see that this theory
does not prove the replacement axiom.



I couldn't understand a word of what you said. Frankly speaking from
the general outlay of it , I don't think it can disprove replacement.

Anyhow does this apply to ZFC+ (ZFC+ is the same as ZFC++ but without
the ur-elements and in which regularity is the same as in ZFC).

I just thought that replacement would be a theorum here, because for
each Small Set x (i.e , x is a set in which exist a nother set that
contains it as a member) F(x)={f(y)|yex} , is always subnumerous or
equinumerous to x, It is impossible for F(x) to be supernumerous to x.
Therefore replacement would follow from global comprhenesion.

Anyhow. I don't know why your prove included a set of ur-elements,
would that prove of yours stands for a set which doens't include ur-
elements of the sizes you've mentioned.

I would have appreciate it if you can simplify your prove a little bit
so that I can understand it ( for instance define these sizes you are
refering too, and what is the significance of having this Union of
them as a model, I mean what replacement is disproved because this
union set you are reffering to exist as a model in this theory.


Zuhair





But just in case what I have said is correct,then this theory
would be correctly named ZFC++,since it contain ZFC and proper
classess and ur-elements.

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Relevant Pages

  • Re: ZFC++
    ... Since this theory has axiom of limitation of size, ... then I thought replacement can be proved in this ... class of all ur-elements or Is S is the class of all ur-elements ... They have the same consistency strength. ...
    (sci.math)
  • Re: ZFC++
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    (sci.math)
  • Re: ZFC++
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  • Re: ZFC in another shape.
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  • Re: ZFC in another shape.
    ... the strong version of replacement. ... Actually there is no need to prove union. ... You see we can define Successor from the axiom of ordinal succession. ... Now we can proceed to define infinity: ...
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